Fermi Thermodynamics problems solved - ?· 1 Enrico Fermi, Thermodynamics, 1937 I. Θερµοδυναµικά…

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  • 1

    Enrico Fermi, Thermodynamics, 1937

    I.

    1. 3.12 L 4.01 L 2.34atm. .

    : W = -P V = -2.34 atm 101325 Pa atm-1 (4.01 3.12) 10-3 m3 = -211 J

    2. 30 g 1 m3 18C.

    : ===PRT

    MmPRT

    MmnRTPV

    atm 0.358kPa 36.3m 1

    K 291.15molK J8.31447

    mol g 2g 30

    31 === P

    3. ( ) 0C.

    : ====RTPM

    VmRT

    MmnRTPV ( P = 1 atm)

    311

    1

    mg 1250

    K 273.15molK J 8.31447mol g 28.01348Pa 101325

    =

    =

    kgm0.800

    g 1250m 11 33

    ====m

    Vv

    4. 10 g 2 20C 1 0.3 atm.

    : :

    ( )1

    2122 lnlnln

    12

    1

    2

    1

    2

    1PP

    MmRTPP

    MmRTdP

    PMmRTdP

    PRT

    MmPPdVW

    P

    P

    P

    P

    P

    P

    ===

    ==

    dPPRT

    MmdP

    PVdV

    PRT

    MmV

    T2=

    ==

    : J 917atm 1atm 0.3ln

    mol g 31.9988K 293.15molK J 8.31447g 10

    1

    1

    =

    =

    W

    .

    1. 3.4 103 J 134 J.

    : U = q + W = 134 J 3.4 103 J = 3266 J

    2. 3 mol 5 atm 3 atm 0C;

    : U = 0. U = q + W W = q

  • 2

    === 1

    2

    1

    2 lnln2

    1PPnRTq

    PPnRTPdVW

    P

    P

    J 348053lnK 273.15molK J 8.31447mol 3 1 == q

    3. 1 mol 291 21000 cm3 305 12700 cm3. (V, P) . .

    : P1 P2

    . VnRTP =

    : kPa 115m 0.021

    K 291molK J 8.31447mol 13

    1

    1 =

    =

    P

    kPa 200m 0.0127

    K 305molK J 8.31447mol 13

    1

    1 =

    =

    P

    200

    150

    100

    50

    0

    P (k

    P)

    0.0250.0200.0150.0100.0050.000V (m

    3)

    1

    2

    , , U = Q + W.

    ( ) ( ) J 1307m 0.0210.0127kPa2

    2001152

    312

    212

    1

    =+

    =+

    == WVVPP

    PdVWV

    V

    :

    dVVUdT

    TUdU

    TV

    +

    = . (

    ) CV, 0 .

    CV , U = CV T = n cV T.

    cV = 3/2 R,

    U = 1 mol 1.5 8.31447 J K-1 mol-1 (305 291) K = 175 J

  • 3

    : q = U W q = 175 J 1307 J = -1132 J

    4. 1.35 . 18C. .

    : Poisson: 1

    1

    212

    122

    111

    ==

    VVTTVTVT

    cV = 5/2 R cP = cV + R = 7/2 R, = 7/5 = 1.4

    : ( ) C55K 3281.35K 273.1518 11.42 o==+= T

    .

    1. 1 mol () Carnot 400 300 . 1 L 5 L. .

    : Poisson 1

    1

    1

    212

    122

    111

    ==

    TT

    VVVTVT

    .

    cV = 3/2 R, cP = 5/2 R, = 5/3. , 1.54 L 7.70 L.

    .

    35

    30

    25

    20

    15

    10

    5

    0

    P (a

    tm)

    876543210V (L)

    T = 400 K

    T = 300 K

    Carnot, , .

    .: 2

    1

    2

    1

    TT

    qq

    =

    ==

    1

    2121 1 T

    TqqqW

  • 4

    . ,

    . , A

    B

    VV

    nRTq ln= .

    , kJ 5.35L 1L 5lnK 400molK J 8.31447mol 1 11-1 ==

    q , , q2 = 4.01 kJ W =

    1.34 kJ.

    2. 400C 18C;

    : 568.027340018400

    2

    12

    2

    12

    2

    =+

    =

    =

    == TTT

    qqq

    qW

    3. 1 J -18C, 38C.

    : J 1.2227318

    27338J 121

    212

    2

    1

    2

    1 =++

    === qTT

    qqTT

    qq

    J 0.22J 1J 1.2212 === qqW

    IV.

    1. 1000 g ; ( = 4.2 J g-1 K-1.

    : ( )====== 12 lnln2

    1

    2

    1

    2

    1

    TTmcTdTmc

    TdTC

    dSSTdTC

    TdqdS P

    T

    TP

    T

    T

    PT

    T

    P

    111

    1

    2 K J 1310K 273K 373lnKg J 4.2g 1000ln === S

    TTmcS P

    2. 1.192.1 10 TPV = . 100 L 0.42 J K-1. T V.

    : dVVUdTCdV

    VUdT

    TUdU

    TV

    TV

    +=

    +

    =

    PVST

    VUPdVTdSdU

    TT

    =

    =

    PdVSdTdF = Maxwell VT T

    PVS

    =

    : 2.11.092.11.19 101.110 VTTPVTP

    V

    =

    = .

    :

  • 5

    ( ) += dVVTVTTdTCdU V 2.11.192.11.09 10101.1dVVTdTCdU V

    2.11.1810 +=

    :

    +=

    +

    = dVVSdT

    TC

    dVVSdT

    TSdS

    T

    V

    TV

    dVVTdTTC

    dS V 2.11.09101.1 +=

    3. 78.3C 3580 J g-1. dP / dT .

    : Clausius Clapeyron:

    K 6.24atm 1K kPa 16.3

    K351.45molK J 8.31447Pa 101325mol g 46g J 3580 12211

    11

    2 ==

    =

    =

    dTdP

    RThP

    dTdP

    , 1 atm.

    V.

    1. .

    : : f = c + 2 p. 2 , . c = 2 p = 2, 2 .

    2. ; ( .)

    : H2O, N2, O2, . c = 3. , f = c + 2 p = 3 + 2 1 = 4 ,. 2 , . , f = 3 + 2 2 = 3, .

    3. , , V: 0.924 + 0.0015 t + 0.0000061 t2, t C. 1 C 18C.

    : , q = 0 U = V Q. :

    dTdVTQVQU =

    . , :

    ( ) ( )+=++== cTbQTcTbTadTdQT

    dTdVQTq 22

    q = -1 C 291 K (0.0015+20.000006118) = -0.50 J

  • 6

    VI.

    1. 2 2 K(T) 18C 5900. 1 atm. .

    : Dalton atm 1AA2 ==+ PPP .

    :

    [ ]RTP

    2A2A = [ ] RT

    PAA =

    ( ) [ ][ ]

    === RTPPPRT

    PP

    TK 2A

    A2

    A

    A2

    2 2

    AA

    ( )( )

    ++=

    TPKTK

    RTP 4112A

    ++

    =

    K 291molK J 8.31446Lmol 5900Pa 101325411

    Lmol 59002K 291molK J 8.31447

    11

    1

    1

    11

    AP

    Pa 95081Pa 6244Pa 101325Pa 62442AA

    === PP

    2AP ,

    2AP . :

    %2.3032.0

    2624495081

    26244

    2

    2A

    A

    A

    2

    ==+

    =+

    = xPP

    P

    x

    2. 1 -210 kJ mol-1, 19C 1 atm.

    : , 19C Gibbs-Helmholz:

    2, T

    HTG

    T nP

    =

    .

    ( ) ( ) ( ) === 200

    0 lnlnlnRTH

    dTTKd

    RTGTKTKRTG

    ( ) ( )( ) ( ) ( )

    =

    =

    =

    21

    0

    1221

    0

    1

    22

    0 11exp11lnlnTTR

    HTKTKTTR

    HTKTK

    TdT

    RHTKd

    ( ) Lmol 4383K 292

    1K 291

    1molK J 8.31447

    mol J 102.1-expLmol 590019 111-15

    1

    =

    =CK o

  • 7

    ++

    =

    K 292molK J 8.31446Lmol 4383Pa 101325411

    Lmol 43832K 292molK J 8.31447

    11

    1

    1

    11

    AP

    Pa 94105Pa 7220Pa 101325Pa 72202AA

    === PP

    %7.3037.0

    2722094105

    27220

    2

    2A

    A

    A

    2

    ==+

    =+

    = xPP

    P

    x

    VII.

    1. 30 g aCl .

    :

    ( )1

    1 L mol 1.027L 12

    mol g 35.4523.00g 301

    =+===

    VMm

    VnC

    atm 25.1bar 25.4Pa 102.54K 298mol K J 8.314m10mol 1.027 611

    33 =====

    RTVnP

    = 298 .

    T = Kf C = 1.84 K L/mol 1.027 mol/L=1.900 K, T = Ke C = 0.512 K L/mol 1.027 mol/L = 0.526 K.

    2. (C6H12O6) NaCl . .

    : RTMmP

    1

    11 = , RTM

    mP2

    22 2= = 21 PP

    = RTMmRT

    Mm

    2

    2

    1

    1 2 17.64527.3598977.22

    9994.15600794.1120107.126222

    1

    2

    1 =+

    ++==

    MM

    mm

    3. .

    : 2 2 . : f = c + 2 p= 2 + 2 2 = 2. .

    4. ( ) . . ( . Clapeyron.)

  • 8

    VIII.

    1. 4000 10torr. ( , .)

    : [ ][ ] [ ]+

    +

    +=

    NaNaNax

    VIII 209:

    310400026000

    23

    926000

    23

    92

    cm mol 1012310400010931010931

    ===

    ..T.x

    xn T (1),

    VNn = Na.

    RTnRTVN

    P ii

    i ==

    .

    ni = n (1-x), Na+ ni = n x . ( ) ( )xnRTxxxnRTP +=++= 11 ( ) 311 m mol 0.0401cmHg 76

    Pa 101325K 4000molJK 8.314

    cmHg 11 ===+

    RTPxn . (2)

    (1) (2) n

    0880m mol 0.0401

    cm mol 103.123

    310.x ==

    2. Debye 3R/2. ( .)

    : (195):

    ( )

    =TRDTC 3 ,

    21

    =

    TD .

    D (196):

    ( )1

    3

    112 1

    1

    0

    33

    =

    eedxxD x

    , D, . = 0.2485 D = 0.5. T = 0. 2485 .

    31/1/2006 10/12/2008

    1.0

    0.8

    0.6

    0.4

    0.2

    0.0

    D

    1.00.80.60.40.20.0T/